I have done most of the legwork, but I have fallen short at the final hurdle. Could you please correct my mistake(s)?

enter image description here

Given the condition $x>0$ I need to consider the infinitely repeated game where the above strategic form is the stage game. The discount factor: $\delta=\frac{1}{2}$.

I need to find a further condition on the game, such that we have a subgame perfect equilibrium in which both players "cooperate" in each period.

What I have done so far

The profile of strictly dominated strategies $(u,r)$ constitutes the cooperative strategy where $x>0$ and $x \ne 1$.

The prescription (cooperative strategy) will therefore be $(u,r)=(x+1,x+1)$

The grim trigger is $(d,l) = (x,x)$

Payoff from obeying the prescription:

$(x+1) + \delta(x+1) + \delta^2(x+1) +\delta^3(x+1)+... = \frac{x+1}{1-\delta}$

Payoff from deviating (here comes the mistake):

$2x + \delta x + \delta^2 x +\delta^3 x +... =\color{red}{\frac{2x-\delta(x+1)}{1-\delta}}$

The part in red should apparently be:


I don't understand why I am wrong, could you please explain? My understanding was that if a player deviates in period 1 in order to obtain $2x$ then both players would have to play the grim trigger profile $(d,l)$ from period 2 to perpetuity. As a result, the players would be missing out on $\delta(x+1)$. The payoff numerator would therefore be $\color{red}{2x - \delta(x+1)}$ instead of $\color{blue}{(2-\delta)x}$.

Given that the correct numerator is $\color{blue}{(2-\delta)x}$, I can easily finish the question:

$$\frac{x+1}{1-\delta} \ge \frac{(2-\delta)x}{1-\delta}$$

Therefore, we would have a subgame perfect equilibrium iff $x \le 2$.

The problem is that my mistake lies in:

$2x + \delta x + \delta^2 x +\delta^3 x +...$ $=\color{red}{\frac{2x-\delta(x+1)}{1-\delta}}$

But, I'm not sure why. I would like to know why I am wrong and why $\color{blue}{(2-\delta)x}$ is the correct numerator.



1 Answer 1


the payoff from play the trigger strategy will be: $$ \sum_{i=0}^{\infty}(x+1) \delta^i=\frac{x+1}{1-\delta}$$

if I deviate and I play $l$ or $d$ the payoff will be $$ 2x + \sum_{i=1}^{\infty}x \delta^i = 2x+ x\frac{\delta}{1-\delta}=\frac{2x(1-\delta) + x\delta}{1-\delta}= \frac{x(2-\delta)}{1-\delta}$$

then, the condition is $$ \frac{x+1}{1-\delta} \geq \frac{x(2-\delta)}{1-\delta}$$

I hope that this will help you ;)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.